INSTRUCTIONS TO THE CANDIDATES
- This paper contains two sections; Section 1 and Section 11.
- Answer all the questions in section 1 and only five questions from Section 11
- All workings and answers must be written.
- Marks may be given for correct working even if the answer is wrong.
- Calculations and KNEC Mathematical tables may be used EXCEPT where stated otherwise.
- Show all the steps in your calculations, giving your answers at each stage.
QUESTIONS
SECTION I (50 MARKS)
Answer all questions in this section in the spaces provided.
- Without using a calculator ,evaluate
Giving your answer as mixed fraction (3mks) - Two boys and a girl shared some money. The younger boy got 5/18 of it; the elder boy got 7/12 of the remainder and the girl got the rest. Find the percentage share of the younger boy to the girl’s share. (4mks)
- Three numbers, 1400,1960 and n have a G.C.Dand L.C.M of 70 and 22 x 52 x 72 x 11 respectively.Find the least possible value of n (3mks)
- A bus starts off from Kitale at 9a.m and travels towards Kakamega at a speed of 60km/hr. At 9.50a.m, a matatu leaves Kakamega and travels towards Kitale at a speed of 60Km/h. How far from Kitale will the two vehicles meet? ( 3mks)
- Find the equation of a straight line which is equidistant from the points A(2,3) and B (6,1) (3mks)
- Simplify the expression completely (3mks)
- Given that sin θ =2/3 and θ is an acute angle, find without using tables tan2 θ + Cos2 θ. Give your answer as a mixed fraction. (3mks)
- Solve for y in the equation below. (4mks)
8(22)y = 6 (2y) – 1 - Using a ruler, a pair of compasses only and (proportional) a set square, construct on the upper side division of line BC, a line BD such that ∠DBC = 37.50. Use the line BD to divide BC into 4 equal portions. (3mks)
- Sketch the net of the solid below. (2mks)
- In a regular polygon, each interior angle is x0 and each exterior angle is
- Find angle X0 (1mk)
- Find the number of sides of the polygon (2mks)
- The figure below represents a plot of land ABCD such that AB= 85m, BC 75m CD= 60m DA = 50m and angle ACB = 900. (not drawn to scale)
Determine the area of the plot, in hectares correct to two decimal places. (4mks) - An open rectangular box measures externally 32cm long, 27cm wide and 15cm deep. The box is made up of metal 1cm thick. If it has a mass of 1.5kg, what is the density of the box to 4 significant figures? (3mks)
- Find the integral values of x which satisfy the following inequalities;
2x + 3 > 5x – 3 > -8 (3mks) - A Kenyan bank buys and sells foreign currency as shown below.
Buying Ksh Selling Ksh 1 US dollar ($) 63.00 63.20 1 UK pound (£) 125.00 125.95
The histogram shown below represents the distribution of marks obtained in attest. The bar marked A has a height of 3.2 units while B has aheight 1.2 units. If the frequency of the class represented by B is 6, find the frequency of the bar represented by A. (3mks )
SECTION II (50 MARKS)
Answer any five questions in this sections in the spaces provided.
- The figure below (not drawn to scale) shows a quadrilateral ABCD inscribed in a circle. AB = 5cm, BC = 8cm, CD = 7cm and AD = 8cm. AC is one of the diagonals of length 10cm.
- Find the size of angle ABC. (3mks)
- Find the radius of the circle. (2mks)
- Hence, calculate the area of the shaded region. (5mks)
- In the figure below → → → →
OB = b , OC = 3OB and OA = a
→ → → → → →- Given that OD = 1/3 OA and AN = ½ AC, CD and AB meet at M. Determine in terms of a and b
→- AB (1mk)
→ - CD (1mk)
- AB (1mk)
- → → → →
Given that CM = k CD and AM = h AB determine the values of the scalars k and h (5mks) - Show that O,M and N are collinear. (3mks)
- Given that OD = 1/3 OA and AN = ½ AC, CD and AB meet at M. Determine in terms of a and b
- Three variables p, q and r are such that p varies directly as q and inversely as the square of r.
- When p = 18, q = 24 and r = 4.
Find p when q = 30 and r = 10. (4mks) - Express q in terms of p and r. (1mk)
- If p is increased by 20% and r is decreased by 10% find:
- A simplified expression for the change in q in terms of p and r. (3mks)
- The percentage change in q. (3mks)
- When p = 18, q = 24 and r = 4.
- A circular lawn is surrounded by a path of uniform width of 7m. The area of the path is 21% that of the lawn.
- Calculate the radius of the lawn. (4 marks)
- Given further that the path surrounding the lawn is fenced on both sides by barbed wire on posts at intervals of 10 metres and 11 metres on the inner and outer sides respectively. Calculate the total number of posts required for the fence. (4 marks)
- Calculate the total cost of the posts if one post costs sh 105. (2 marks)
- The diagram below represents two vertical watch – towers AB and CD on a level ground. P and Qare two points on a straight road BD. The height of the tower AB is 20m and road BD is 200m
- A car moves from B towards D. At point P, the angle of depression of the car from pointsA is 11.30Calculate the distance BP to 4 significant figures. (2mks)
- If the car takes 5 second to move from P to Q at an average speed of 36km/hr, calculate the angle of depression of Q from A to 2 decimal places. (3mks)
- Given that QC = 50.9 cm, calculate
- The height of CD in meters to 2 decimal places; (2mks )
- The angle of elevation of A from C to the nearest degree. ( 3mks)
- The parents of acertain mixed secondary school decided to buy a school van worth Ksh. 900,000. Each student was to contribute the same amount of money. 50 students were transferred from the school; as a result each f the remaining students had to pay Ksh. 600 more.
- Find the original number of the students in the school. (5mks)
- Find the percentage change in contributions per student. (3mks)
- If the ratio of boys to girl in the school was 11:7 find the amount money contributed by boys alone.(2mks)
- Five members of ‘SILK’, a self-supporting enterprise Jane, Jepchoge, Esther, Mama Charo and Chepkoech were given a certain amount of money to share amongst themselves. Jane got 3/8 of the total amount while Jepchoge got 2/5 of the remainder. The remaining amount was shared equally among Esther, Mama Charo and Chepkoech each of which received Kshs. 6,000;
- How much was shared among the five business women? (3mks)
- How much did Jepchoge get? (2mks)
- Jane, Jepchoge and Chepkoech invested their money and earned a profit of Kshs. 12,000. A third of the profit was left to maintain the business and the rest was shared according to their investments. Find how much each got. (5mks)
- Onyango and Juma live 200km apart. One day, Onyango left his house at 7.00am and travelled towards Juma’s house at an average speed of 30km/hr. Juma left his house at 7.30am on the same day and travelled towards Onyango’s at an average speed of 40km/hr.
- Determine:-
- The time they met. (2mks)
- The distance from Onyango’s house where the two met. (2mks)
- How far was Onyango from Juma’s house when they met? (2mks)
- The two took 15 minutes at the meeting point and then travelled to Juma’s house at an average speed of 20km/hr. Find the time he arrived at Juma’s house.(2mks)
- Determine:-
MARKING SCHEME
SECTION 1
- 3/4 + 15/7 ÷ 4/7 × 7/3
= 3 + 9 = 21 + 36 = 57
4 7 28 28
= (80 − 35)× 2
56 3
=45/56 × 2/3
=15/28
=57/28 × 28/15 = 19/5
= 34/5 - 7/12 × 13/18 =91/216
= 91 × 5 = 91 + 60 = 151
216 18 216 216
(5/18 ÷ 216/65)100
= 5 × 216 × 100 = 1200
18 65 13
= 924/13% -
1400 2
2
2
5
5
7700
350
175
35
7
1
1960 2
2
2
5
7
7980
490
245
47
7
1
N = 23 × 5 × 72
70 2
535
7
23 × 52 × 72 × 11
n = 2× 5 × 11 × 7
= 770 -
At 9.50am , the bus has travelled
(20/60 × 60) = 20km
The distance between the two vehicles at 9.50am
(65 − 45) =20km
Rel . speed = 140 km/ h.
It will take them 45/140hrs since leaving Kitale.
Distance covered at the time Matatu met the bus
=45/140 × 80 = 25.71km -
Gradient = 1−3 = −1
6−2 2
g × −1/2 =−1
g=2
y−2 =2
x−4
y−2=2x−8
y=2x −6 - 4x(3x−4) = 4x(3x−4)
20−15x+4x−3x2 5(4−3x) + x(4−3x)
= 4x(3x−4)
(5+x)(4−3x)
= −4x(4−3x)
(5+x)(4−3x)
= −4x
5+x -
y=√9−4
=√5
tan2 θ + Cos2 θ
( /√5)2 + ( √5/3)2
= 4 + 5 = 61
5 9 45
=116/45 - 8x(22)y = 6(2y)−1
=8x(2y)2 = 6x(2y)−1
Let 2y = x
∴ 8x2 = 6x−1
→8x2−6x+1=0
→8x2−4x−2x+1=0
4x(2x−1)−1(2x−1)=0
→(4x−1)(2x−1)=0
→∴x = ½ or ¼
2y=½ = 2−1 or 2y =¼ = 2−2
∴y=−1
or
y=−2 -
-
- x + x−36 =1800
3
→3x + x−36 =540
4x =576
→x= 1440
Exterior angle= 144 − 36 =108 =360
3 3
∴No of sides of the polygon
n= 360 =10sides
36 - AC2= 852 − 752
AC ==40m
Area of quadrilateral ABCD
½ × 40 × 75 +
=1500 + √984375
=2492m2
In hectares = 2492 = 0.2492hectares
10000
≈0.25hectares -
External volume =(32 × 27 × 15) =12960cm3
Internal volume =(30 × 25 × 14) =10500cm3
Volume of material = (12960 − 10500)
Metre used = 2460cm3
Density = mass = 1500
volume 2460
=0.6098g/cm3 - 2x>5x−3>−8
2x+3>5x−3
−3x>6
x<2
5x−3>−8
5x>−5
x>−1
−1<x<2
The integral values are 0 and 1 -
Remaining amount = 9600 × 95 × 125
1E 100
=Ksh 1,140,000
Amount speed = ksh 1,140,000 × 3
4
The Balance = ksh 1,140,000- 855,000
=ksh285,000
Amount in dollars = 285,000
63.20
=4509.49 USdollars - 1.2 × 10 × x = 6
x = 6/12
=0.5
3.2 × 15 × 0.5 = 24
SECTION II
-
102 =82 + 52 − 2 × 8 × 5 Cos B
Cos B=89 − 100 = −11
80 80
B = Cos−1(−11/80)=97.900- 2R = 10
Sin 97.90
R = 5
Sin 97.90
=5.0479cm - Sin A = 7/10 Sin 82.1 = 0.6984
A = Sin−1(0.6934) = 43.900
∠COD = 2 × 43.900= 87.800
Area =87.80 × 22 × 5.0479 − ½ ×5.04329 Sin 87.80
360 7
= 19.5316 − 12.7313
= 6.800cm2
-
-
- →
AB = b−a - →
CD = −3b + 1/3a
- →
- CM = 1/3K a − 3b
→AM = −h a + h b
CM = CA+AM
=1/3K a − 3 b
= a(1− h) + b(h−3)
K=3−3h and 1/3K =1−h
h=¾
k=¾ - →
OM=¼a + ¾b
→
ON =½a + ¾b
→ →
OM = ½ON
Hence parallel, and point O is common
→ O, M and N are collinear
-
-
- p ∞ q/r2
p =kq/r2
18 =24 × k
42
K= 18 × 16
2
K=12
Equation: p=12q/r2
when q=30, r=10
P=(12 × 30)
100
= 18/5 or 3.6 - P= 12q/r2
12q= pr2
q = pr2/12 -
- New value of P =1.2P
New r = 0.9r
q =pr2/12
=1.2p × (0.9r)2
12
Δin q =
= pr2(1.2 × 0.81 – 1)
12
=–0.028 pr2
12 - –0.028 pr2 × 12 × 100
12 pr
=2.8%
- New value of P =1.2P
- p ∞ q/r2
-
- R = r + 7
πR2 − πr2 = 21 πr2
100
πR2=0.21 πr2 + πr2
πR2 = 1.21r2
R2 = 1.21r2
R =1.2r
R + 7 = 1.1r
7= 1.1r−r
7= 0.1r
R=70m - Inner radius =70m
2πr = 2 × 22× 70 =400m
7
440m = 44 Posts
10
Outer radius r=77
2πr = 2 × 22× 77 =484m
7
484 = 44posts
11
Total number of posts =88 - Total cost
88 × 105 = Ksh 9240
- R = r + 7
-
- Tan 11.30 =20
BP
∴BP = 20 = 100.09m
Tan 11.30 - Time taken in hrs = 5
3600
Distance netween P and Q
=
=0.05km
=50km
→ tan α = 20
150.09
α = tan−1 ( 20 ) = 7.590
(150.09) -
- CD2 = 50.92 − 49.912 = 99.8019
CD = √99.8019 =9.9900
= 9.99m - tan α = 10.01 = 0.05005
200
α = tan−1(0.05005)
= 2.865
=3.00
- CD2 = 50.92 − 49.912 = 99.8019
- Tan 11.30 =20
- 900,000 − 900,000 = 600
x − 50 x
1500 − 1500 =1
x −50 x
1500x − 1500x + 75000
x(x − 50)
75000 =x2 − 50x
x2 − 50x −75000 = 0
x2 −300x + 250x −75000 =0
x(x − 300) + 250(x − 300) =0
(x − 300)(x + 250) =0
→x=300 - 900,000 − 900,000 × 100%
250 300
= (3600 − 3000) × 100%
( 3000 )
= 20% - B : G
11:7
11 × 900,000
18
=Kshs 550,000
- 900,000 − 900,000 = 600
-
- Let the constant amount be x
Jane-(3/8x)/=
Jepchoge's- 2/5(5/8x)/=
=¼x/=
Remaining 3/5x − 18,000
x=48,000/=
Therefore the original amount is 48,000 - Jepchoge received
(¼ × 48,000)
= 12,000/= - Business maintenance
=(1/3 × 12,000)
= 4,000/=
Balance = 8,000/=
Ratios: Jane=(3/8 × 48,000)
= Kshs. 18,000
Jepchoge=(¼ × 48,000)
=Kshs. 12,000
Chepkoech=1/3 × 18,000)
=Kshs. 6,000
Ratio: 18,000 : 12,000 : 6,000
3 : 2 : 1
Jane got 3/6 × 8,000
=Kshs. 4,000
Jepchoge got 2/6 × 8,000
=Kshs. 2,667
Chepkoech got 1/6 × 8,000
=Kshs.1,333
- Let the constant amount be x
-
-
R.S = 30km/hr + 40km/hr
= 70km/hr √m1
D= S×T
= 30km/hr ×½hr
=15km
T = D
R.S
= 185km
70km/hr
= 2 ½ hours √m1
Time they met = 7.30am √m1
+2.30
= 10.00am √m1-
Distance from Onyango’s house where the two met:
Distance = 30km/hr x 5/2 hours
= 75km
= 75 + 15
= 90km √m1 -
Distance from Onyango to Juma’s house when they met:
200km – 90km √m1
= 110km √A1
-
Meeting time = 10.00a.m
+ 15
10.15am
T = 110km = 5 ½ hours√m1
20km/hr
Time of arrival to Jumas house
= 10.15am
+5.30√m1
= 15.45pm or 3.45pm
-
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